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3 years ago

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A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it

's position?
A. 68.6 J
B. 137 J
C. 274 J
D. 960 J

1 answer:

Alexeev081 [22]3 years ago

3 0

It should be B.137 just took the test.

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Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

3 years ago

The battery charger for an mp3 player contains a step-down transformer with a turns ratio of 1:38, so that the voltage of 120 v

matrenka [14]

Transformer contains two coils: primary and secondary. They allow change of voltage to lower or higher value. In first case we have step-down and in second case we have step-up transformer.
Formula used for transformer is: $\frac{N_{1} }{N_{2}} = \frac{V_{1}}{V_{2}}$

Where:N1 = number of turns on primary coilN2 = number of turns on secondary coilV1 = voltage on primary coilV2 = voltage on secondary coil
In a step-down transformer primary coil has more turns than secondary coil. So the ratio 1:38 means that for each turn on secondary coil we have 38 turns on primary coil.
We can solve the equation for V2: $V_{2} = \frac{ V_{1}* N_{2} }{ N_{1} } \\ V_{2} = \frac{ 120* x }{ 38x} } \\ V_{2} = 3.16V$

Secondary coil provides voltage of 3.16V.

7 0

4 years ago

A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud

Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride 6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

= 9.28 cos 27.27°

= 8.24 km

horizontal axis component = 9.28 sin ∅

= 9.28 sin 27.27°

= 4.5 km

distance of the final point from the origin

= 7.99 -(6.2-4.5)

= 6.29 km

displacement

$d = \sqrt{6.29^2+8.24^2}$

d = 10.37 km

b) $tan \theta = \dfrac{6.29}{8.24}$

θ = 37.36°

4 0

3 years ago

If you connect two identical storage batteries together in series (" " to "-"), and place them in a circuit, the combination wil

Lelu [443]

Answer:

4 (D): twice the voltage, and the same current will flow through each.

Explanation:

When we have batteries in series in a circuit, the equivalent voltage will be the sum of their voltages, and the current flowing through then will be the same.

These batteries are identical, so they have the same voltage, therefore when they are put in series, the equivalent voltage will be twice the voltage of one battery.

So the correct option is 4 (D): twice the voltage, and the same current will flow through each.

3 0

3 years ago

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.70

Aleksandr-060686 [28]

Given:
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s

Let ω₂ = the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
= (0.7/3.5)*(30.159)
= 6.032 rad/s

ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s

Answer: 0.96 rev/s

3 0

3 years ago